Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am trying to solve a problem in harmonic functions in Rudin's book(Real and Complex analysis 3rd edition)

To clarify the problem I want to ask, we need some notations:

(1) $U$ is the open unit disc, and $T$ is the unit circle, the boundary of $U$ in the complex plane

(2) $P(z,e^{it})$ is the Poisson Kernel. $$P(z,e^{it})=\frac{1-|z|^2}{|e^{it}-z|^2}$$ for $z\in U$, $e^{it}\in T$

(3)$P[f]$ is the Poisson Integral against $f\in L^1(T)$

(4)$P[d\mu]$ is the Poisson Integral against a complex measure on $T$, defined by $$P[d\mu](z)=\int_T P(z,e^{it})d\mu(e^{it})\quad (z\in U)$$

(5)$C(T)$ is the space consisting of all the continuous complex functions on $T$

(6)We associate to any function $u$ in $U$ a family of functions $u_r$ on $T$, defined by $$u_r(e^{it})=u(re^{it})\quad(0\leq r<1)$$

(7)The measure $\sigma$ is defined by $\sigma=m/2\pi$, where $m$ is ordinary Lebesgue measure on $T$

(8)$||u_r||_1$ is defined by $$||u_r||_1=\int_T |u_r|d\sigma\quad(0\leq r<1)$$

The problem is:

Suppose $u$ is harmonic in $U$, and $\{u_r:0\leq r<1\}$ is a uniformly integrable subset of $L^1(T)$. Modify the proof of Theorem 11.30 to show that $u=P[f]$ for some $f\in L^1(T)$.

Before stating Theorem 11.30, one needs theorem 11.29.

Theorem 11.29: Suppose that (a)$X$ is a separable Banach space, (b)${\Lambda_n}$ is a sequence of linear functionals on $X$, (c)$sup_n||\Lambda_n||=M<\infty$

Then there is a subsequence $\{\Lambda_{n_i}\}$ such that the limit $$\Lambda x=\lim_{i\to\infty}\Lambda_{n_i} x$$ exists for every $x\in X$. Moreover, $\Lambda$ is linear, and $||\Lambda||\leq M$

Proof (Sketch): Note that $\{\Lambda_n\}$ is pointwise bounded and equicontinuous. Since each point of $X$ is a compact set, Theorem 11.29 follows from Arzela-Ascoli Theorem. Besides, it is obvious that $||\Lambda||\leq M$ and that $\Lambda$ is linear.

Theorem 11.30: Suppose $u$ is harmonic in $U$, and $$sup_{0<r<1} ||u_r||_1=M<\infty$$ It follows that there is a unique complex Borel measure $\mu$ on $T$ so that $u=P[d\mu]$

Proof: Define linear functionals $\Lambda_r$ on $C(T)$ by $$\Lambda_r g=\int_T gu_rd\sigma\quad (0\leq r<1)$$ Therefore, $||\Lambda_r||\leq M$. By Theorem 11.29 and Riesz representation theorem for the dual of $C(T)$ there is a measure $\mu$ on $T$, with $||\mu$$||\leq M$, and a sequence $r_j\to 1$, so that $$\lim_{j\to\infty}\int_T gu_{r_j}d\sigma=\int_T gd\mu\quad (*)$$ for every $g\in C(T)$.

Put $h_j(z)=u(r_j z)$. Then $h_j$ is harmonic in $U$, continuous on $\bar{U}$, and is therefore the Poisson integral of its restriction to $T$. Fix $z\in U$, and apply $(*)$ with $$g(e^{it})=P(z,e^{it})$$ Since $h_j(e^{it})=u_{r_j}(e^{it})$, we obtain $$u(z)=\lim_j u(r_j z)=\lim_j h_j(z)$$, and $$\lim_j h_j(z)=\lim_j\int_T P(z,e^{it})h_j(e^{it})d\sigma(e^{it})=\int_T P(z,e^{it})d\mu(e^{it})=P[d\mu](z)$$

To prove uniqueness, it suffices to show that $P[d\mu]=0$ implies $\mu=0$.

Pick $f\in C(T)$, put $u=P[f]$, $v=p[d\mu]$. By Fubini's theorem, and the symmetry $P(re^{i\theta},e^{it})=P(re^{it},e^{i\theta})$, $$\int_T u_rd\mu=\int_T v_rfd\sigma\quad (0\leq r<1)$$ When $v=0$ then $v_r=0$, and since $u_r\to f$ uniformly, as $r\to 1$, we conclude that $$\int_T fd\mu=0$$ for every $f\in C(T)$ if $P[d\mu]=0$. By Riesz representation theorem, $\mu=0$.

Any hints will be appreciated. I've really no idea how to modify the proof of Themorem 11.30, because the $L^1$-boundedness of the family $\{u_r\}$ seemes to play an important role in the proof. However, I cannot see the relationship between the boundedness and uniformly integrability. I have goolged this problem, but I cannot find anything helpful.

Again, any hints will be appreciated.

share|improve this question

1 Answer 1

up vote 2 down vote accepted

First,Theorem 11.30 draw to showing that

$$\sup_{0<r<1} ||u_r||_1=M<\infty$$

In fact,for a r meet the conditions above.Due to $u_r$ is a harmonic function,the sets $E_1=u_r^{+-1}(\mathbb{R})$ and $E_2=u_r^{--1}(\mathbb{R})$ are measureable.where $u_r=u_r^{+}-u_r^{-}$ .

split $E_j$ into $E_{ji},\mu(E_{ji})<\delta(j=1,2)$ and use the definition of uniformly integrable we can get there exist $M$ such that $sup_{0<r<1} ||u_r||_1=M<\infty$ for all r.

Second,use the Theorem 11.30,we get a Borel measure $\mu$ and $u=P[\mu]$. Then by Lebesgue decomposition,$d\mu = fd\sigma+d\mu_s$ where $f\in L^1(\mu)$ .

Finally,use the definition of uniformly integrable again, it's clear that $\mu_s=0$.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.